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\begin{document}

\title{高等代数一}
\subtitle{2-排列的逆序数、行列式的第二种定义}
%\institute{上海立信会计金融学院}
%\author{王立庆}
\author{ {\ppr LQW} }
\renewcommand{\today}{\number\year \,年 \number\month \,月 \number\day \,日}
%\date{2020年10月15日}
\date{{ \ppr 2022年9月22日} }
\maketitle

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\begin{frame}{目录 }

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\begin{enumerate}

\item 什么是排列的逆序数
\item 按指标的排列的逆序数来定义行列式的值
\item 证明行列式的二种定义的结果是相等的
\item 证明行列式的转置不改变行列式的值

\end{enumerate}


\end{frame}

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\begin{frame}{2.1. 例子1：上三角行列式}

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\begin{itemize}
\item  一个行列式 $|a_{ij}|_{n\times n}$ 的{\color{red}对角线}元素是指 $a_{11},a_{22},\cdots,a_{nn}$ 这 $n$ 个元素。
\item  {\color{red}上三角行列式}是指对角线的左下方的元素都为零的行列式。
\item  例子1：上三角行列式的值等于它的对角线的元素的乘积。例如，
\begin{eqnarray*}
\begin{vmatrix} a_{11}&a_{12}&a_{13} \\ 0 &a_{22}&a_{23} \\ 0 & 0 &a_{33} \\  \end{vmatrix}
= a_{11}a_{22}a_{33}. 
\end{eqnarray*}

\item 证明：两次使用行列式的{\color{red}按第一列展开}的定义，我们有
\begin{eqnarray*}
\begin{vmatrix} a_{11}&a_{12}&a_{13} \\ 0 &a_{22}&a_{23} \\ 0 & 0 &a_{33} \\  \end{vmatrix}
=a_{11}\begin{vmatrix} a_{22}&a_{23} \\  0 &a_{33} \\  \end{vmatrix}
= a_{11}a_{22}a_{33}. 
\end{eqnarray*}

\end{itemize}

\end{frame}

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\begin{frame}{2.2. 例子2}

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\begin{itemize}
\item  例子2：按第一列展开，计算下述行列式的值，
\begin{eqnarray*}
D=\begin{vmatrix} 
a_{11}&a_{12}&a_{13} &a_{14} \\ 
0 &a_{22}&a_{23} &a_{24} \\ 
0 & 0 &a_{33} &a_{34} \\  
0 & 0 &a_{43} &a_{44} \\  
\end{vmatrix}. 
\end{eqnarray*}

\item 解答：三次使用行列式的{\color{red}按第一列展开}的定义，我们有
\begin{eqnarray*}
D = a_{11} \begin{vmatrix} 
a_{22}&a_{23} &a_{24} \\ 
0 &a_{33} &a_{34} \\  
0 &a_{43} &a_{44} \\  
\end{vmatrix}
= a_{11}a_{22}\begin{vmatrix} 
a_{33} &a_{34} \\  
a_{43} &a_{44} \\  
\end{vmatrix}
%=a_{11}a_{22} \left( a_{33}a_{44} - a_{34}a_{43} \right) 
= a_{11}a_{22}a_{33}a_{44} - a_{11}a_{22}a_{43}a_{34}. 
\end{eqnarray*}

\end{itemize}

\end{frame}

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\begin{frame}{2.3. 例子3}

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\begin{itemize}
\item  例子3：按第一列展开，计算下述行列式的值，
{\footnotesize
\begin{eqnarray*}
D=\begin{vmatrix} 
a_{11}&a_{12}&a_{13} &a_{14} \\ 
a_{21} &a_{22}&a_{23} &a_{24} \\ 
0 & 0 &a_{33} &a_{34} \\  
0 & 0 &a_{43} &a_{44} \\  
\end{vmatrix}. 
\end{eqnarray*}
}

\item 解答：三次使用行列式的{\color{red}按第一列展开}的定义，我们有
{\footnotesize
\begin{eqnarray*}
D &=& a_{11} \begin{vmatrix} 
a_{22}&a_{23} &a_{24} \\ 
0 &a_{33} &a_{34} \\  
0 &a_{43} &a_{44} \\  
\end{vmatrix}
- a_{21} \begin{vmatrix} 
a_{12}&a_{13} &a_{14} \\ 
0 &a_{33} &a_{34} \\  
0 &a_{43} &a_{44} \\  
\end{vmatrix} 
= a_{11}a_{22} \begin{vmatrix} a_{33} &a_{34} \\  a_{43} &a_{44} \\  \end{vmatrix}
- a_{21} a_{12} \begin{vmatrix} a_{33} &a_{34} \\  a_{43} &a_{44} \\  \end{vmatrix} \\ 
&=& a_{11}a_{22} \left( a_{33}a_{44} - a_{43}a_{34} \right) 
- a_{21}a_{12} \left( a_{33}a_{44} - a_{43}a_{34} \right).  
\end{eqnarray*}
}

\end{itemize}

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\begin{frame}{2.4. 行列式的按指标排列的定义}

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\begin{itemize}

\item {\color{red}定义：行列式 $D=|a_{ij}|_{n\times n}$ 的值定义为下述和式
\begin{eqnarray*}
D = \begin{vmatrix} 
a_{11} & a_{12} & \cdots & a_{1n} \\    
a_{21} & a_{22} & \cdots & a_{2n} \\    
\cdots & \cdots & \cdots & \cdots  \\
a_{n1} & a_{n2} & \cdots & a_{nn} \\    
\end{vmatrix}
 = \sum\limits_{ \sigma=(j_1,j_2,\cdots,j_n)\in S_n } (-1)^{\pi(\sigma)} a_{1j_1}a_{2j_2}\cdots a_{nj_n}. 
\end{eqnarray*}
其中的符号解释如下：}
\begin{enumerate}
\item  {\color{red}$\sigma=(j_1,j_2,\cdots,j_n)$ 是数字 $1,2,\cdots,n$ 的一个全排列。}
\item  {\color{red}$S_n$ 是数字 $1,2,\cdots,n$ 的所有全排列组成的集合。}
\item  {\color{red}$\pi(\sigma)$ 是全排列 $\sigma$ 的逆序数，即前大后小的数对的个数。}
\end{enumerate}

\end{itemize}

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\begin{frame}{2.5. 例子4 }

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\begin{itemize}

\item  例子4：五阶行列式 $D=|a_{ij}|_{5\times 5}$ 写出来一共有 $5!=120$ 项，求其中一项
 $$a_{{\color{red}1}5}a_{{\color{red}2}4}a_{{\color{red}3}3}a_{{\color{red}4}1}a_{{\color{red}5}2}$$ 前面的正负号。

\item  解答：

\begin{enumerate}

\item  第一个指标（行指标）已经按 ${\color{red}12345}$ 排列。

\item  第二个指标（列指标）对应的全排列是 $\sigma=54312$, 其中前大后小的数对为 $54,53,51,52,43,41,42,31,32$, 共9对，所以
\begin{eqnarray*}
\pi(\sigma) = \pi(54312) = 9. 
\end{eqnarray*}

\item  所以 $a_{15}a_{24}a_{33}a_{41}a_{52}$ 这一项前面的正负号是 $(-1)^9=-1$. 

\end{enumerate}

\end{itemize}

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\begin{frame}{2.6.  }

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\begin{itemize}

\item 五阶行列式的 120 项的其中一项，
{\footnotesize 
\begin{eqnarray*}
D = \begin{vmatrix} 
a_{11}&a_{12}&a_{13}&a_{14}& \underline{\color{red}a_{15}} \\ 
a_{21} &a_{22}&a_{23}&\underline{\color{red}a_{24}}&a_{25} \\ 
a_{31} &a_{32}&\underline{\color{red}a_{33}}&a_{34}&a_{35} \\ 
\underline{\color{red}a_{41}} &a_{42}&a_{43}&a_{44}&a_{45} \\ 
a_{51} &\underline{\color{red}a_{52}}&a_{53}&a_{54}&a_{55} \\ 
\end{vmatrix}
= \cdots + (-1)^9 {\color{red}a_{15}a_{24}a_{33}a_{41}a_{52}} +\cdots . 
\end{eqnarray*}
}

\end{itemize}

\end{frame}


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\begin{frame}{2.7. 证明：行列式的两种定义是一致的（以四阶为例） }

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\begin{enumerate}\itemsep0em

\item 按第一列展开的方式来定义行列式，
{\footnotesize
\begin{eqnarray*}
\begin{vmatrix} 
{\color{red}a_{11}} &a_{12}&a_{13}&a_{14} \\ 
{\color{red}a_{21}} &a_{22}&a_{23}&a_{24} \\ 
{\color{red}a_{31}} &a_{32}&a_{33}&a_{34} \\ 
{\color{red}a_{41}} &a_{42}&a_{43}&a_{44} 
\end{vmatrix}
= {\color{red}a_{11}} A_{11} + {\color{red}a_{21}} A_{21} + {\color{red}a_{31}} A_{31} + {\color{red}a_{41}} A_{41}. 
\end{eqnarray*}
}

\item  按指标排列的方式来定义行列式，（这里先将列指标按1234排列）
{\footnotesize 
\begin{eqnarray*}
\begin{vmatrix} 
{\color{red}a_{11}} &a_{12}&a_{13}&a_{14} \\ 
{\color{red}a_{21}} &a_{22}&a_{23}&a_{24} \\ 
{\color{red}a_{31}} &a_{32}&a_{33}&a_{34} \\ 
{\color{red}a_{41}} &a_{42}&a_{43}&a_{44} 
\end{vmatrix}
=\sum\limits_{ \sigma=(i_1,i_2,i_3,i_4)\in S_4 } (-1)^{\pi(\sigma)} {\color{red}a_{i_11}} a_{i_22}a_{i_33}a_{i_44}. 
\end{eqnarray*}
}

\item 比较可知，形如 $a_{i_11} a_{i_22}a_{i_33}a_{i_44}$ 的24项一一对应，而且正负号也一样。

\end{enumerate}

\end{frame}

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\begin{frame}{2.8. 例子5}

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\begin{itemize}
\item  例子5：按行列式的按指标排列的定义，写出行列式代表的数学表达式，
{\footnotesize 
\begin{eqnarray*}
D=\begin{vmatrix} a&0&0&b \\ 0&c&d&0 \\ 0&e&f&0 \\ g&0&0&h \end{vmatrix}. 
\end{eqnarray*}
}

\item  解答：按每行选一个数、四行选出四个数、同时使得这四个数落在不同的列，可得 $acfh, adeh, bcfg, bdeg$ 共四种。
因为我们这里是一行行往下选，所以行指标的次序总是1234. 
{\footnotesize  
\begin{eqnarray*}
D &=& (-1)^{\pi(1234)} acfh + (-1)^{\pi(1324)} adeh + (-1)^{\pi(4231)} bcfg +(-1)^{\pi(4321)} bdeg \\ 
&=& (-1)^{0} acfh + (-1)^{1} adeh + (-1)^{5} bcfg +(-1)^{6} bdeg \\
&=& acfh - adeh - bcfg + bdeg. 
\end{eqnarray*}
}

\end{itemize}

\end{frame}

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\begin{frame}{2.9. 行列式的转置}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{itemize}
\item  {\color{red}定义：从行列式 $|a_{ij}|_{n\times n}$ 出发，将每个 $(i,j)$ 位置的元素与 $(j,i)$ 位置的元素对换，得到的行列式称为原行列式的转置。}

\item  例如，将行列式 $D$ 进行转置，得到行列式 $D'$, 
\begin{eqnarray*}
D = \begin{vmatrix} a&b&c \\ u&v&w \\ x& y&z \end{vmatrix}
\hspace{0.3cm}
\longrightarrow 
\hspace{0.3cm}
D' = \begin{vmatrix} a&u&x \\ b&v&y \\ c&w&z \end{vmatrix}. 
\end{eqnarray*}

\item  定理：转置不改变行列式的值。即有 $D=D'$. 

%\item  例子：
%\begin{eqnarray*}
%\begin{vmatrix} a&b \\ c&d \end{vmatrix} = ad-bc 
%= \begin{vmatrix} a&c \\ b&d \end{vmatrix}. 
%\end{eqnarray*}

\end{itemize}

\end{frame}

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\begin{frame}{2.10. 证明：转置不改变行列式的值 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
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\begin{enumerate}

%\item 定理：转置不改变行列式的值。证明如下。
%\begin{myitemize}
\item 对二阶行列式，直接验证 
%\begin{eqnarray*}
{\footnotesize 
$\begin{vmatrix} a&b \\ c&d \end{vmatrix}
= \begin{vmatrix} a&c \\ b&b \end{vmatrix}. 
$
}%\end{eqnarray*}
等式两边都是 $ad-bc$. 
\item 对三阶行列式，仍然可以直接验证。
\item 对 $n$ 阶行列式，我们有下述等式，
{\footnotesize 
\begin{eqnarray*}
 (-1)^{\pi(j_1,j_2,\cdots, j_n)} a_{1j_1}a_{2j_2}\cdots a_{nj_n} 
&=&  (-1)^{\pi(1,2,\cdots,n)+\pi(j_1,j_2,\cdots, j_n)}  a_{1j_1}a_{2j_2}\cdots a_{nj_n}  \\
&=& (-1)^{\pi(2,1,\cdots,n)+\pi(j_2,j_1,\cdots, j_n)}  a_{2j_2}a_{1j_1}\cdots a_{nj_n}  \\
&=& \cdots \cdots \\ 
&=& (-1)^{\pi(i_1,i_2\cdots,i_n)+\pi(1,2,\cdots, n)}  a_{i_11}a_{i_22}\cdots a_{i_nn},
\end{eqnarray*}
}
%其中 $a_{1j_1}a_{2j_2}\cdots a_{nj_n} = a_{i_11}a_{i_22}\cdots a_{i_nn}$, 
同样的 $n$ 个元素，开始是按照行指标从小到大排列，每次交换相邻的两个元素，最后变成按照列指标从小到大排列。

\item {\color{red}行指标的逆序数和列指标的逆序数的和，在每次交换两个元素的过程中，奇偶性是不变的。}

\end{enumerate}

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\begin{itemize}

\item  例子6：确定五阶行列式的展开表达式中的一项前面的正负号，
{\footnotesize 
\begin{eqnarray*}
D = \begin{vmatrix} 
a_{11}&a_{12}&a_{13}&\underline{\color{red}a_{14}}& a_{15} \\ 
\underline{\color{red}a_{21}} &a_{22}&a_{23}&a_{24}&a_{25} \\ 
a_{31} &a_{32}&a_{33}&a_{34}&\underline{\color{red}a_{35}} \\ 
a_{41} &a_{42}&\underline{\color{red}a_{43}}&a_{44}&a_{45} \\ 
a_{51} &\underline{\color{red}a_{52}}&a_{53}&a_{54}&a_{55} \\ 
\end{vmatrix}. 
\end{eqnarray*}
}

\end{itemize}

\end{frame}

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\begin{frame}{2.12. 例子6的解答 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
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\begin{enumerate}

\item  按照行指标从小到大来写，$a_{14}a_{21}a_{35}a_{43}a_{52}$  前面的正负号是 
$$(-1)^{\pi (41532)} = (-1)^6 =+1.$$

\item  按照列指标从小到大来写，$a_{21}a_{52}a_{43}a_{14}a_{35}$  前面的正负号是 
$$(-1)^{\pi (25413)} = (-1)^6 =+1.$$

\item  按照看到哪写到哪，$a_{21}a_{14}a_{35}a_{43}a_{52}$  前面的正负号是 
$$(-1)^{\pi(21345)+\pi(14532)} = (-1)^{1+5}=+1. $$

\end{enumerate}

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\begin{frame}{2.13. 课堂练习 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{enumerate}

\item  六阶行列式 $D=|a_{ij}|_{6\times 6}$ 写出来一共有 $6!=720$ 项，求其中一项
 $$a_{31}a_{24}a_{53}a_{15}a_{62}a_{46}$$ 前面的正负号。
 
\item  按指标的排列的逆序数来确定正负号的方式，计算下述行列式的值，
{\footnotesize
\begin{eqnarray*}
D = \begin{vmatrix} 
0&2&0&0&0&0 \\ 
3&0&2&0&0&0 \\ 
0&3&0&2&0&0 \\ 
0&0&3&0&2&0 \\ 
0&0&0&3&0&2 \\ 
0&0&0&0&3&0 \\ 
\end{vmatrix}. 
\end{eqnarray*}
}


\end{enumerate}

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\begin{enumerate}

\item  正号。

\item  $D=-216$. 

\end{enumerate}

\end{frame}

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\end{document}


